question_answer

A voltmeter having least count 0.1 V and an ammeter having least count 0.2 A are used to measure the potential difference across the ends of a wire and current flowing through the wire respectively. If the reading of voltmeter is 4.4 V and reading of ammeter is 2.2 A, then find (i) the resistance of wire with maximum permissible error and (ii) maximum percentage error.

Answer:

                Here, V = 4.4V, I = 2.2A                                                       \[\Delta \text{V}=0.\text{1V},\,\,\Delta \text{I}=0.\text{2A}\]                                                                                 \[\text{R }=\frac{V}{I}=\frac{4.4}{2.2}=2\,ohm\] Maximum permissible error                                            \[\frac{\Delta R}{R}=\frac{\Delta V}{V}+\frac{\Delta I}{I}=\frac{0.1}{4.4}+\frac{0.2}{2.2}\]                                                                         \[=0.0\text{23}+0.0\text{91}=0.\text{114}\]                            \[\Delta \text{R}=0.\text{114}\times \text{R}=0.\text{114}\times \text{2}=0.\text{228}\]                                                                                     \[\Delta \text{R}=0.\text{2W}\]  (rounding off to one decimal place) Resistance of wire with max. permissible error R = (2.0+ 0.2) ohm Max. percentage error =                           \[\frac{\Delta R}{R}\times \text{1}00=0.\text{114}\times \text{1}00=\text{11}.\text{4}%\]